Question

$A(3,7)$, $B(5,11)$, $C(-2,8)$ are the vertices of a $\Delta ABC$, $AD$ is one of the median of triangle. Find the equation of median $AD$.

Answer 1

Given : $A(3,7)$ ; $B(5,11)$ , $C(-2,8)$ are vertices of $\Delta ABC$
$AD$ is one of the medians
Find : Equation of median $AD$
$\therefore AD$ is one of the medians
$\therefore D$ is the midpoint of $BC$
$D(x,y)=(\frac{5-2}{2},\frac{11+8}{2})$ ...... (midpoint formula)
$D(x,y)=(\frac{3}{2},\frac{19}{2})$
Let $A(3,7)=(x_1,y_1)$
$D(\frac{3}{2},\frac{19}{2})=(x_2,y_2)$
By two-point form, $\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}$
$\Rightarrow \frac{x-3}{3-\frac{3}{2}}=\frac{y-7}{7-\frac{19}{2}}$
$\Rightarrow \frac{x-3}{\frac{6-3}{2}}=\frac{y-7}{\frac{14-19}{2}}$
$\Rightarrow \frac{x-3}{\frac{3}{2}}=\frac{y-7}{\frac{-5}{2}}$
$\Rightarrow \frac{x-3}{3}=\frac{y-7}{-5}$ ....(Cancelling $2$ on both sides)
$\Rightarrow -5x+15=3y-21$
$\Rightarrow -5x-3y+15+21=0$
$\Rightarrow -5x-3y+36=0$
$\Rightarrow 5x+3y-36=0$
$\therefore$ equation of median $AD=5x+3y-36=0$.